Question

How many unpaired electrons are present in each of the following?

\(\begin{aligned}{}(a){\left( {Co{F_6}} \right)^{3 - }}(highspin)\\(b){\left( {Mn{{(CN)}_6}} \right)^{3 - }}(lowspin)\\(c){\left( {Mn{{(CN)}_6}} \right)^{4 - }}(lowspin)\\(d){\left( {MnC{l_6}} \right)^{4 - }}(highspin)\\(e){\left( {RhC{l_6}} \right)^{3 - }}(lowspin)\end{aligned}\)

Short answer

a) It has \(4\) unpaired electrons.

b) It has \(2\) unpaired electrons.

c) It has \(1\) unpaired electron.

d) It has \(5\) unpaired electrons.

e) It has \(0\) unpaired electrons

Step-by-step solution

of 1: Explanation

(a)

• In\({\left( {{\rm{Co}}{{\rm{F}}_6}} \right)^{3 - }}\), the central atom is the transition metal with 6 fluoride ions attached in it.
• Now, let us determine the oxidation state of \(Co\)as we use \(x\)for the oxidation state of \(Co\).
• As we know the general oxidation state of fluoride ion is \( - 1,\) we can determine the oxidation state of \(Co\)like this:

\(\begin{aligned}{}x + (6 \cdot - 1) &= - 3\\x - 6 &= - 3\\x = - 3 + 6\\x &= + 3\end{aligned}\)

• It means the oxidation state of \(Co\)is \( + 3.\)
• As per the periodic table, the atomic number of \(Co\) is \(27\) and its electronic configuration is \((Ar)3{d^7}4{s^2}\).
• Hence, In oxidation state of \(C{o^{3 + }}\)the electronic configuration is: \((Ar)3{d^4}4{s^0}\)
• The molecule has a high spin and it means the electron pairs donated by ligand go to the outermost orbitals.
• From the crystal field diagram we can see that \({\left( {{\rm{Co}}{{\rm{F}}_6}} \right)^{3 - }}\)has \(4\)unpaired electrons.

(b)

• In \({\left( {Mn{{(CN)}_6}} \right)^{3 - }}\), the central atom is the transition metal and it is attached with 6 cyanide ions.
• Let us now determine the oxidation state of \(Mn\)as we use \(x\)for the oxidation state of\(Mn\)
• As we know that general oxidation state of cyanide ion is \( - 1,\)we can determine the oxidation state of \(Mn\)like this:

\(\begin{aligned}{}x + (6 \cdot - 1) &= - 3\\x - 6 &= - 3\\x &= - 3 + 6\\x &= + 3\end{aligned}\)

• It means the oxidation state of \(Mn\)is \( + 3.\)
• The atomic number of \(Mn\)is \(25\) and its electronic configuration is \((Ar)3{d^5}4{s^2}\).
• In oxidation state of \(M{n^{3 + }}\)the electronic configuration is: \((Ar)3{d^4}4{s^0}\)
• The molecule has a low spin and it means the electron pairs donated by ligand go to the innermost orbitals.
• From the crystal field diagram we can see that \({\left( {Mn{{(CN)}_6}} \right)^{3 - }}\)has \(2\) unpaired electrons.

(c)

• In\({\left( {Mn{{(CN)}_6}} \right)^{4 - }}\), the central atom is the transition metal which is attached with 6 cyanide ions.
• Now, let us determine the oxidation state of \(Mn\)as we use \(x\)for the oxidation state of \(Mn\).
• As we are aware that the general oxidation state of cyanide ion is \( - 1,\)so we can determine the oxidation state of \(Mn\)like this:

\(\begin{aligned}{}x + (6 \cdot - 1) &= - 4\\x - 6 &= - 4\\x& = - 4 + 6\\x &= + 2\end{aligned}\)

• It means the oxidation state of \(Mn\)is \( + 2\)
• If we look at the periodic table we can see that the atomic number of \(Mn\)is \(25\)and its electronic configuration is \((Ar)3{d^5}4{s^2}\)
• In oxidation state of \(M{n^{2 + }}\)the electronic configuration is: \((Ar)3{d^5}4{s^0}\)
• The molecule has a low spin, which means the electron pairs donated by ligand go to the innermost orbitals.
• From the crystal field diagram it is clear that \({\left( {Mn{{(CN)}_6}} \right)^{4 - }}\)has \(1\)unpaired electron.

(d)

• In\({\left( {{\rm{MnCl}}{l_6}} \right)^{4 - }}\), the central atom is the transition metal which is attached with 6 chloride ions.
• Let us now determine the oxidation state of \(Mn\)as we use \(x\)for the oxidation state of
• As we are aware that general oxidation state of chloride ion is \( - 1,\)so we can determine the oxidation state of \(Mn\)like this:
• \(\begin{aligned}{}x + (6 \cdot - 1) &= - 4\\x - 6 &= - 4\\x &= - 4 + 6\\x &= + 2\end{aligned}\)
• It means the oxidation state of \(Mn\)is \( + 2\)
• The atomic number of \(Mn\)is \(25\)and its electronic configuration is \((Ar)3{d^5}4{s^2}\).
• In oxidation state of \(M{n^{2 + }}\)the electronic configuration is: \((Ar)3{d^5}4{s^0}.\)
• The molecule has a high spin, which means the electron pairs donated by ligand go to the outermost orbitals.
• From the crystal field diagram we can see that \({\left( {MnC{l_6}} \right)^{4 - }}\)has \(5\)unpaired electrons.

(e)

• In\({\left( {RhC{l_6}} \right)^{3 - }}\), the central atom is the transition metal which is attached with 6 chloride ions.
• Let us determine the oxidation state of \(Rh\)as we use \(x\)for the oxidation state of \(Rh\).
• As we know that general oxidation state of chloride ion is \( - 1,\)so we can determine the oxidation state of \(Rh\)like this:

\(\begin{aligned}{l}x + (6 \cdot - 1) &= - 3\\x - 6 & = - 3\\x& = - 3 + 6\\x &= + 3\end{aligned}\)

• It means the oxidation state of \(Rh\)is\( + 3\).
• The atomic number of \(Rh\)is \(45\)and its electronic configuration is \((Kr)4{d^7}5{s^2}\).
• In oxidation state of \(R{h^{3 + }}\)the electronic configuration is: \((Kr)4{d^6}5{s^0}.\)
• The molecule has a low spin, which means the electron pairs donated by ligand go to the innermost oritals.
• From the crystal field diagram we can see that \({\left( {RhC{l_6}} \right)^{3 - }}\)has \(0\)unpaired electrons.

Result

a) It has \(4\) unpaired electrons.

b) It has \(2\) unpaired electrons.

c) It has \(1\) unpaired electron.

d) It has \(5\) unpaired electrons.

e) It has \(0\) unpaired electrons.